Can You Solve This Physics Problem on Kinetic and Potential Energy?

[riseofphoenix]

c, d, and e I don't know how to solve

For d I tried doing

v_f² = v_i² +2ad
V_f = √(2*9.8*2.3) = 6.71 m/s

Assuming the inclined plane is frictionless
Since m2 is roped to m1, it's velocity up the ramp will be 6.71m/s too

The total increase in KE is then 0.5[8+6]*6.71^2 = 315.16 J

^

"INCORRECT"

Right now I need a genius to tell me how to get the answer ASAP -.-

Since (d) is wrong, so is e. -.-
Story of my life.

 

[berkeman]

c, d, and e I don't know how to solve

For d I tried doing

v_f² = v_i² +2ad
V_f = √(2*9.8*2.3) = 6.71 m/s

Assuming the inclined plane is frictionless
Since m2 is roped to m1, it's velocity up the ramp will be 6.71m/s too

The total increase in KE is then 0.5[8+6]*6.71^2 = 315.16 J

^

"INCORRECT"

Right now I need a genius to tell me how to get the answer ASAP -.-

Since (d) is wrong, so is e. -.-
Story of my life.

For (c), how did you get that number? Isn't it just the sum of the two ΔU values?

 

[riseofphoenix]

For (c), how did you get that number? Isn't it just the sum of the two ΔU values?

m1 falls 2.3m and therefore has -6*9.8*2.3 = -135.24 J difference.

m2's PE increases 8*9.8*sin 30° = +39.2 J difference.

The total change then is 39.2 - 135.24 = -96.04 J

^ That was my previous answers, but I got that wrong too...
So I'm pretty much screwed.

 

[berkeman]

m1 falls 2.3m and therefore has -6*9.8*2.3 = -135.24 J difference.

m2's PE increases 8*9.8*sin 30° = +39.2 J difference.

The total change then is 39.2 - 135.24 = -96.04 J

^ That was my previous answers, but I got that wrong too...
So I'm pretty much screwed.

You left off the 2.3 for m2...

 

[riseofphoenix]

You left off the 2.3 for m2...

(c)

m1 falls 2.3m and therefore has -6*9.8*2.3 = -135.24 J difference.

m2's PE increases 8*9.8*sin 30°*2.3 = +90.16 J difference.

The total change then is 90.16 - 135.24 = -45.08 J

Ok so how does this relate to part d and e though? How would I solve for those two last ones?

(d)

(e)

 

[berkeman]

(c)

m1 falls 2.3m and therefore has -6*9.8*2.3 = -135.24 J difference.

m2's PE increases 8*9.8*sin 30°*2.3 = +90.16 J difference.

The total change then is 90.16 - 135.24 = -45.08 J

Ok so how does this relate to part d and e though? How would I solve for those two last ones?

(d)

(e)

(d) asks for the change in KE. Since the system is frictionless...

 

[riseofphoenix]

(d) asks for the change in KE. Since the system is frictionless...

But I already tried solving for it and i got it wrong... (??)

 

[berkeman]

But I already tried solving for it and i got it wrong... (??)

But you had the wrong answer for (c) before. Now you have that right.

What is the total energy of a system in terms of the potential and kinetic energies? If the system is lossless (no friction), what can you say about the total energy over time...?

 

[riseofphoenix]

But you had the wrong answer for (c) before. Now you have that right.

What is the total energy of a system in terms of the potential and kinetic energies? If the system is lossless (no friction), what can you say about the total energy over time...?

KE_initial + PE_initial = KE_final + PE_final
KE_initial + PE_initial = 0 + 0 (No friction)
KE_initial + PE_initial = 0?

 

[berkeman]

KE_initial + PE_initial = KE_final + PE_final
KE_initial + PE_initial = 0 + 0 (No friction)
KE_initial + PE_initial = 0?

Rearrange so you show the relationship between ΔKE and ΔPE...

 

[riseofphoenix]

KE = -PE
KE = -mv?

 

[berkeman]

KE = -PE
KE = -mv?

If you put Δs in front of your first equation, you would be on the right track...

 

[Perpendicular]

Consider the total energy over time which is a constant. Hence change in PE must balance out change in KE. The rest is obvious.

 

[Perpendicular]

As for (e) consider that you have a final change in KE, as well as the masses. You can now plug that into solve for velocity.

 

[berkeman]

Oops, Looks like riseofphoenix is on a temporary vacation from the PF for a different thread.